1. t-test - paired Testing for a difference

2. What does it do? Tests for a difference in means Compares two cases (eg soil moisture content north & south of a hedge) Data must occur in natural pairs  Measurements at matched points on either side of a hedge  Measurements at corresponding points on two shores  Two measurements on the same person

3. If your data are not normal, but are paired, use Wilcoxon If your data are normal, but are not paired, use unpaired t-test Your data occur in natural pairs You have five or more pairs of values Your data are normally distributed  Only continuous data (lengths, weights etc) can be normally distributed  You can do a check by seeing if your data look roughly like this: Planning to use it? Make sure that…

4. How does it work? You assume (null hypothesis) there is no difference between the two means The test works with the differences of the pairs of values. If, for example, the values in the first sample were all much bigger, then all the differences would be positive The mean and standard deviation of the differences is calculated, and put into a formula.

5. Doing the test These are the stages in doing the test: 1.Write down your hypotheseshypotheses 2.Finding the differencesdifferences 3.Finding the mean & standard deviationmean & standard deviation 4.Use the formula to calculate the t-valueformula 5.Write down the degrees of freedomdegrees of freedom 6.Look at the tablestables 7.Make a decisiondecision Click here Click here for an example

6. Hypotheses H 0: mean1 = mean2 For H 1, you have a choice, depending on what alternative you were looking for. H 1: mean1 > mean2 eg: Soil moisture content is higher on the north of the hedge than the south of the hedge orH 1: mean1  mean2 eg: Soil moisture content is different on the north and south sides of the hedge. Unless you have a good scientific reason for expecting one to be larger, you should choose “different” for H 1

7. Differences Work out the differences between the pairs of values, including signs Always do sample 1 – sample 2 not the other way round, where sample 1 is the one you expect to have larger values

8. Mean & Standard Deviation To find the mean of your differences, add them all up and divide by how many there are To find the standard deviation s, use the formula n = number of values  x 2 = total of the squares of the differences These calculations can be done on a spreadsheet or graphic calculator

9. Formula Substitute your values into the formula = mean of the differences s = standard deviation of the differences n = number of values

10. Degrees of freedom The formula here for degrees of freedom is degrees of freedom = n – 1 Where n is the number of values You do not need to worry about what this means –just make sure you know the formula! But in case you’re interested – the fewer values you have, the more likely you are to get a large t-value by chance – so the higher your value has to be to be significant.

11. Tables This is a t table Degrees of freedom Significance levels - note different values for 1 and 2-tailed

12. Make a decision If your value is bigger than the tables value then you can reject the null hypothesis. Otherwise you must accept it. Make sure you choose the right tables value – it depends whether your test is 1 or 2 tailed:  If you are using H 1 : mean1 > mean 2, you are doing a 1-tailed test  If you are using H 1: mean1  mean 2, you are doing a 2-tailed test

13. Example: Soil Moisture North & South of Hedge Data were obtained for soil moisture content at seven matched points north and south of a hedge. Hypotheses: H 0: Mean moisture content on north = mean moisture content on south. H 1 Mean moisture content on north > mean moisture content on south

14. The data Site1 2 34 5 6 7 North 4.092.933.8810.503.505.144.63 South 4.65 6.00 3.47 4.33 2.20 2.50 3.33

15. Differences We are expecting the North side to have a higher moisture content. So we do North - South N4.09 2.93 3.88 10.50 3.50 5.14 4.63 S 4.65 6.00 3.47 4.33 2.20 2.50 3.33 N – S-0.56-3.070.41 6.17 1.30 2.64 1.30

16. Mean & Standard Deviation Our data (the differences) are: -0.56-3.070.41 6.17 1.30 2.64 1.30  x 2 = (-0.56) 2 + (-3.07) 2 + … + 2.64 2 + 1.30 2 = 58.3251

17. The test Degrees of freedom = 7 – 1 = 6 We are doing a 1-tailed test Tables value (5%) = 1.943 Since our value is smaller than the tables value, we accept H 0 – the moisture content on the north is not significantly higher.